Find the Fallacy

Before starting reading this note please go through my note on uncountability of RR. There I have shown that RR, the set of real numbers, is uncountable. Now in this note, I will show a that there exists a bijection between RR and NN. Well, obviously such bijection doesn't exists and my method has a fallacy, but I want you people to find out the fallacy yourself.

I will construct a function f:NRf:N\rightarrow R as follows:

From NN choose an arbitrary element n1{ n }_{ 1 } and map it to an arbitrary element x1{ x }_{ 1 } from RR.

Next choose an arbitrary element n2{ n }_{ 2 } from N{n1}N-\{ { n }_{ 1 }\} and map it to arbitrary element x2{ x }_{ 2 } of R{x1}R-\{ { x }_{ 1 }\} and continue the process.

Thus in the i-th step map the arbitrary element niN{n1,n2,...,ni1}{ n }_{ i }\in N-\{ { n }_{ 1 },{ n }_{ 2 },...,{ n }_{ i-1 }\} to an arbitrary element xiR{x1,x2,...,xi1}{ x }_{ i }\in R-\{ { x }_{ 1 },{ x }_{ 2 },...,{ x }_{ i-1 }\} .

Then, from the construction of ff, ff is injective. Choose any two different elements from NN, and observe that they are mapped to two elements of RR under ff, which must be different (different because whenever in a step an element of NN is mapped to that of RR, the latter is deleted from RR before proceeding to the next step). Again ff is surjective because choose any arbitrary element from RR, see that it is the image of an element of NN (Since our process of construction of ff is infinite, this means at some step of our process we must have mapped an element of NN to that element of RR).

So ff turns out to be bijective. Clearly, there is a strong fallacy in my considerations. Can anybody point it out?

Note by Kuldeep Guha Mazumder
3 years, 9 months ago

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Well, you need to ensure thatff is indeed a function. Where have you done that?

User 170039 - 3 years, 8 months ago

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Every natural number is mapped to exactly one real number. That's enough to prove that it is a function.

Ivan Koswara - 3 years, 8 months ago

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Consider the same argument with the position of N\mathbb{N} and R\mathbb{R} reversed. Will you say that in this case also ff (if we retain the same notation) is a function?

User 170039 - 3 years, 8 months ago

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@User 170039 Yes, of course.

EDIT: Actually, ff won't be a function, exactly because R>N|\mathbb{R}| > |\mathbb{N}|. First, because R\mathbb{R} is uncountable, so the process won't finish even with infinite time. Second, since R>N|\mathbb{R}| > |\mathbb{N}|, at some point there's no more element of N\mathbb{N} that hasn't been paired with.; the process of making ff is halted in this way. If we allow to pick any element of N\mathbb{N} and we can handle the "and continue arbitrarily" portion, then yes, ff becomes a function.

Ivan Koswara - 3 years, 8 months ago

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@Ivan Koswara Yes, precisely that is the point.

User 170039 - 3 years, 8 months ago

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Since our process of construction of ff is infinite, this means at some step of our process we must have mapped an element of RR to that element of NN

This doesn't follow. Just consider the simple function f:NNf : \mathbb{N} \to \mathbb{N} where f(n)=n+1f(n) = n+1. The process is infinite, yet 1 (or 0, if your natural numbers start with 0) has no pre-image.

Ivan Koswara - 3 years, 9 months ago

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Yes that's exactly the fallacy. The fallacy lies in the fact that the method by which the above function is being constructed doesn't guarantee surjectivity. Assume it to be surjective, then consider a closed subset of RR and apply the logic that I used while proving the uncountability of RR.

Kuldeep Guha Mazumder - 3 years, 9 months ago

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